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A238655
Number of partitions of n having standard deviation σ > 3.
4
0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 5, 10, 15, 23, 33, 49, 68, 99, 133, 179, 246, 334, 432, 583, 756, 974, 1261, 1618, 2076, 2657, 3336, 4228, 5270, 6592, 8190, 10182, 12567, 15533, 19008, 23307, 28410, 34622, 42041, 50959, 61487, 74259, 88734, 106666, 127587
OFFSET
1,10
COMMENTS
Regarding "standard deviation" see Comments at A238616.
EXAMPLE
There are 30 partitions of 9, whose standard deviations are given by these approximations: 0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 1.
MATHEMATICA
z = 53; g[n_] := g[n] = IntegerPartitions[n]; c[t_] := c[t] = Length[t];
s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, c[t]}]/c[t]]
Table[Count[g[n], p_ /; s[p] > 3], {n, z}] (*A238655*)
Table[Count[g[n], p_ /; s[p] > 4], {n, z}] (*A238656*)
Table[Count[g[n], p_ /; s[p] > 5], {n, z}] (*A238657*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st dev's of partitions of 30*)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 03 2014
STATUS
approved